Question

我正在使用select distinct以便仅返回不同的数据，但我希望它基于所有列的区别而不包括区别中的agg_source_tag_tag_name。

这是我的问题：

SELECT DISTINCT agg_article_title,agg_article_link,agg_article_media,agg_article_description,agg_article_source_name, agg_source_tag_tag_name, agg_source_url
FROM agg_article join agg_source ON agg_article_source_name = agg_source_name join agg_source_tag ON agg_source_name = agg_source_tag_source_name 
WHERE agg_source_included = 1

我得到以下结果：

agg_article_title | agg_article_link | agg_article_media | agg_article_description | agg_article_source_name | agg_source_tag_tag_name | agg_source_url |
---------------------------------------------------------------------------------------------------------------------------------------------------------
some title 1      | some link 1      | some media 1      | some description 1      | some source name 1      | tag1                    | someurl1       |
---------------------------------------------------------------------------------------------------------------------------------------------------------
some title 1      | some link 1      | some media 1      | some description 1      | some source name 1      | tag2                    | someurl1       |

我得到以下内容返回，因为所有列实际上都是因为agg_source_tag_tag_name而不同，但是，我只想要返回一行，因为我想从区别中省略agg_source_tag_tag_name。

Answer 1

尝试使用WITH GROUP BY

  SELECT agg_article_title,agg_article_link,agg_article_media,agg_article_description,agg_article_source_name, agg_source_tag_tag_name, agg_source_url
  FROM agg_article 
  join agg_source 
  ON agg_article_source_name = agg_source_name 
  join agg_source_tag 
  ON agg_source_name = agg_source_tag_source_name 
  WHERE agg_source_included = 1 
  GROUP BY agg_article_title, agg_article_link

你可以根据自己想要的东西进行分组。

从所有列中选择不同的数据，但是1

1 个答案: